Question: If $x \veebar y = 3x-8y$ and $x \star y = x^{2}-4y^{2}$, find $(-6 \star -2) \veebar 3$.
First, find $-6 \star -2$ $ -6 \star -2 = (-6)^{2}-4(-2)^{2}$ $ \hphantom{-6 \star -2} = 20$ Now, find $20 \veebar 3$ $ 20 \veebar 3 = (3)(20)-(8)(3)$ $ \hphantom{20 \veebar 3} = 36$.